Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 554

\[\boxed{\mathbf{554}\mathbf{.}}\]

\[1)\ \frac{\sqrt{3} \bullet \left( \cos{75{^\circ}} - \cos{15{^\circ}} \right)}{1 - 2\sin^{2}{15{^\circ}}} =\]

\[= \frac{- 2\sqrt{3} \bullet \sin\frac{90{^\circ}}{2} \bullet \sin\frac{60{^\circ}}{2}}{\cos^{2}{15{^\circ}} - \sin^{2}{15{^\circ}}} =\]

\[= \frac{- 2\sqrt{3} \bullet \sin{45{^\circ}} \bullet \sin{30{^\circ}}}{\cos{30{^\circ}}} =\]

\[= \frac{- 2\sqrt{3} \bullet \frac{\sqrt{2}}{2} \bullet \frac{1}{2}}{\frac{\sqrt{3}}{2}} =\]

\[= - \frac{\sqrt{6}}{2} \bullet \frac{2}{\sqrt{3}} = - \sqrt{2}\]

\[2)\ \frac{2\cos^{2}\frac{\pi}{8} - 1}{1 + 8\sin^{2}\frac{\pi}{8} \bullet \cos^{2}\frac{\pi}{8}} =\]

\[= \frac{2\cos^{2}\frac{\pi}{8} - \left( \cos^{2}\frac{\pi}{8} + \sin^{2}\frac{\pi}{8} \right)}{1 + 2 \bullet \left( 2\sin\frac{\pi}{8} \bullet \cos\frac{\pi}{8} \right)^{2}} =\]

\[= \frac{\cos^{2}\frac{\pi}{8} - \sin^{2}\frac{\pi}{8}}{1 + 2\sin^{2}\frac{\pi}{4}} =\]

\[= \frac{\cos\frac{\pi}{4}}{1 + 1 - \cos\frac{\pi}{2}} = \frac{\cos\frac{\pi}{4}}{2 - \cos\frac{\pi}{2}} =\]

\[= \frac{\frac{\sqrt{2}}{2}}{2 - 0} = \frac{\sqrt{2}}{4}\]